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Exercises 10.1.2 Instructor Use Only K. Campbell Solutions

Not For Distribution To Students

10.1 Exercises

Convert the angles into the DMS system. Round each of your answers to the nearest second. 1∘ = 60′, 1′ = 60′′

1. 63.75∘ = 63∘ + 0.75∘ = 63∘ + (0.75∘) (60′

1∘ ) = 63∘ + 45′ = 63∘ 45′

2. 200.325∘ = 200∘ + 0.325∘ = 200∘ + (0.325∘) (60′

1∘ ) = 200∘ + 19.5′

= 200∘ + 19′ + 0.5′ = 200∘ + 19′ + (0.5′) (60′′

1′ ) = 200∘ + 19′ + 30′′

= 200∘ 19′ 30′′

3. −317.06∘ = −317∘ − 0.06∘ = −317∘ − (0.06∘) (60′

1∘ ) = −317∘ − 3.6′

= −317∘ − 3′ − 0.6′ = −317∘ − 3′ − (0.6′) (60′′

1′ ) = −317∘ − 3′ − 36′′

= −317∘ 3′ 36′′

4. 179.999∘ = 179∘ + 0.999∘ = 179∘ + (0.999∘) (60′

1∘ ) = 179∘ + 59.94′

= 179∘ + 59′ + 0.94′ = 179∘ + 59′ + (0.94′) (60′′

1′ ) = 179∘ + 59′ + 56.4′′

≈ 179∘ 59′ 56′ Convert the angles into decimal degrees. Round each of your answers to three decimal places.

1∘ = 60′, 1′ = 60′′

5. 125∘ 50′ = 125∘ + 50′ = 125∘ + (50′) (1∘

60′) = 125∘ + (50

60)

∘≈ 125.833∘

6. −32∘ 10′ 12′′ = −32∘ − 10′ − 12′′ = −32∘ − (10′) (1∘

60′) − (12′′) (1∘

60′) (1′

60′′)

= −32∘ − (10

60)

∘

− (12

3600)

∘

= −32.17∘

7. 502∘ 35′ = 502∘ + 35′ = 502∘ + (35′) (1∘

60′) = 502∘ + (35

60)

∘≈ 502.583∘

8. 237∘ 58′ 43′′ = 237∘ + 58′ + 43′′ = 237∘ + (58′) (1∘

60′) + (43′′) (1∘

60′) (1′

60′′)

= 237∘ + (58

60)

∘

+ (43

3600)

∘

≈ 237.979∘

Convert the angle from degree measure into radian measure, giving the exact value in terms of 𝜋.

360∘ = 2𝜋 radians ⟹ 1∘ =𝜋

180 radians

29. 0∘ = 0 (𝜋

180) radians = 0 radians 30. 240∘ = 240 (

𝜋

180) radians =

4𝜋

3 radians

31. 135∘ = 135 (𝜋

180) radians =

3𝜋

4 radians 32. −270∘ = −270 (

𝜋

180) radians = −

3𝜋

2 radians

33. −315∘ = −315 (𝜋

180) radians = −

7𝜋

4 radians 34. 150∘ = 150 (

𝜋

180) radians =

5𝜋

6 radians

35. 45∘ = 45 (𝜋

180) radians =

𝜋

4 radians 36. −225∘ = −225 (

𝜋

180) radians = −

5𝜋

4 radians

Convert the angle from radian measure into degree measure.

360∘ = 2𝜋 radians ⟹ 1 radian = (180

𝜋)

∘

37. 𝜋 radians = 𝜋 (180

𝜋)

∘= 180∘ 38. −

2𝜋

3 radians = (−

2𝜋

3) (

180

𝜋)

∘= −120∘



39. 7𝜋

6 radians = (

7𝜋

6) (

180

𝜋)

∘= 210∘ 40.

11𝜋

6radians = (

11𝜋

6) (

180

𝜋)

∘= 330∘

41. 𝜋

3radians = (

𝜋

3) (

180

𝜋)

∘= 60∘ 42.

5𝜋

3radians = (

5𝜋

3) (

180

𝜋)

∘= 300∘

43. −𝜋

6radians = (−

𝜋

6) (

180

𝜋)

∘= −30∘ 44.

𝜋

2radians = (

𝜋

2) (

180

𝜋)

∘= 90∘

The average speed of an object equals the distance traveled divided by the elapsed time. For motion along a circle, we distinguish between linear speed and angular speed. Suppose that an object moves around a circle of radius 𝑟 at a constant speed. If 𝑠 is the distance traveled in time 𝑡 around this circle, then the linear speed 𝑣 of the

object is defined as 𝑣 =𝑠

𝑡

As this object travels around the circle, suppose that 𝜃 (measured in radians) is the central angle swept out in time 𝑡 The angular speed 𝜔 of this object is the angle 𝜃 (measured in radians) swept out,

divided by the elapsed time 𝑡, 𝜔 =𝜃

𝑡

Equation 10.2. Velocity for Circular Motion: For an object moving on a circular path of radius 𝑟 with constant angular velocity 𝜔, the (linear) velocity of the object is given by 𝑣 = 𝑟𝜔.

50. A yo-yo, which is 2.25 in. in diameter, spins at a rate of 4500 rev/min. How fast is the edge of the yo-yo spinning in mph? Round your answer to two decimal places.

𝜔 = (4500 revolutions

1 minute) (

2𝜋 radians

1 revolution) (

60 minutes

1 hour) = 540,000𝜋

radians

hour

𝑣 = 𝑟𝜔 =1

2(2.25 inches) (

540,000𝜋

1 hour) (

1 foot

12 inches) (

1 mile

5280 feet) ≈ 30.12

mi

hr

51. How many rev/min would the yo-yo in exercise 50 have to complete if the edge of the yo-yo is to be spinning at a rate of 42 mph? Round your answer to two decimal places.

42 =1

2(2.25 in) (

1 ft

12 in) (

1 mi

5280 ft) (

𝑥 rev

1 min) (

2𝜋 rad

1 rev) (

60 min

1 hr)

⟹ 𝑥 = 42 (2

2.25) (

12

1) (

5280

1) (

1

2𝜋) (

1

60)

rev

min≈ 6274.52

rev

min

52. In the yo-yo trick ‘Around the World,’ the performer throws the yo-yo so it sweeps out a vertical circle whose radius is the yo-yo string. If the yo-yo string is 28 in. long and the yo-yo takes 3 seconds to complete one revolution of the circle, compute the speed of the yo-yo in mph. Round your answer to two decimal places.

𝜔 = (1 rev

3 sec) (

60 sec

1 min) (

60 min

1 hr) (

2𝜋 rad

1 rev) = 2400𝜋

rad

hr

𝑣 = 𝑟𝜔 = (28 in) (1 ft

12 in) (

1 mi

5280 ft) (

2400𝜋 rad

1 hr) ≈ 3.33

mi

hr

53. A computer hard drive contains a circular disk with diameter 2.5 in. and spins at a rate of 7200 RPM. Find the linear speed of a point on the edge of the disk in mph.

𝜔 = (7200 rev

1 min) (

60 min

1 hr) (

2𝜋 rad

1 rev) = 864,000𝜋

rad

hr

𝑣 = 𝑟𝜔 =1

2(2.5 in) (

1 ft

12 in) (

1 mi

5280 ft) (

864,000𝜋 rad

1 hr) ≈ 53.55

mi

hr



54. A rock got stuck in the tread of my tire and when I was driving 70 mph, the rock came loose and hit the inside of the wheel well of my car. How fast, in mph, was the rock traveling when it came out of the tread? (The tire has a diameter of 23 in.) linear speed is given – no computation necessary – 70 mph

55. The Giant Wheel at Cedar Point is a circle with diameter 128 feet which sits on an 8 foot tall platform making its overall height 136 feet. It completes two revolutions in 2 minutes and 7 seconds. Assuming the riders are at the edge of the circle, how fast are they traveling in mph?

𝜔 = (2 rev

127 sec) (

2𝜋 rad

1 rev) (

60 sec

1min) (

60 min

1 hr) =

14,400𝜋 rad

127 hr

𝑣 = 𝑟𝜔 =1

2(128 ft) (

1 mi

5280 ft) (

14,400 𝜋 rad

127 hr) ≈ 4.32

mi

hr

56. Consider the circle of radius 𝑟 pictured below with central angle 𝜃, measured in radians, and subtended arc length 𝑠. Prove that the area of the

shaded sector is 𝐴 =1

2𝑟2𝜃. Hint: use the proportion

𝐴

area of circle=

𝑠

circumference of circle

𝐴

𝜋𝑟2=

𝑠

2𝜋𝑟⟹ 𝐴 =

𝑠𝜋𝑟2

2𝜋𝑟=

1

2𝑠𝑟

𝜃 =𝑠

𝑟⟹ 𝑠 = 𝜃𝑟 ⟹ 𝐴 =

1

2(𝜃𝑟)𝑟 =

1

2𝑟2𝜃

Use the result of exercise 56 to compute the areas of the circular sectors with the given central angles and radii.

Note that the formula requires angles to be in radian measure.

57. 𝜃 =𝜋

6, 𝑟 = 12 ⟹ 𝐴 = (

1

2) (12)2 (

𝜋

6) = 12𝜋 𝑢𝑛𝑖𝑡𝑠2

58. 𝜃 =5𝜋

4, 𝑟 = 100 ⟹ 𝐴 = (

1

2) (100)2 (

5𝜋

4) = 6250𝜋 𝑢𝑛𝑖𝑡𝑠2

59. 𝜃 = 330∘, 𝑟 = 9.3 ⟹1

2(9.3)2(330) (

𝜋

180) =

31713𝜋

400𝑢𝑛𝑖𝑡𝑠2 ≈ 249.07 𝑢𝑛𝑖𝑡𝑠2

60. 𝜃 = 𝜋, 𝑟 = 1 ⟹ 𝐴 =1

2(1)2(𝜋) =

𝜋

2𝑢𝑛𝑖𝑡𝑠2

61. 𝜃 = 240∘, 𝑟 = 5 ⟹ 𝐴 =1

2(5)2(240) (

𝜋

180) =

50𝜋

3𝑢𝑛𝑖𝑡𝑠2

62. 𝜃 = 1∘, 𝑟 = 117 ⟹ 𝐴 =1

2(117)2(1) (

𝜋

180) =

1521𝜋

40𝑢𝑛𝑖𝑡𝑠2 ≈ 119.46 𝑢𝑛𝑖𝑡𝑠2

63. Imagine a rope tied around the Earth at the equator. Show that you need to add only 2𝜋 feet of length to the rope in order to lift it one foot above the ground around the entire equator. (You do not need to know the radius of the Earth to show this.) Let 𝑅 = the radius of the Earth in feet Amount of rope needed: 𝑠 = 𝜃𝑟 = 2𝜋𝑅 To lift 1 foot off the earth: 𝑟 = 𝑅 + 1 is the new radius Amount of rope needed: 𝑠 = 𝜃𝑟 = 2𝜋(𝑅 + 1) Amount of additional rope needed: 2𝜋(𝑅 + 1) − 2𝜋𝑅 = 2𝜋𝑅 + 2𝜋 − 2𝜋𝑅 = 2𝜋



10.2 Exercises

Use the results developed throughout the section to find the requested value. Theorem 10.1. The Pythagorean Identity: sin2(𝜃) + cos2(𝜃) = 1

21. sin(𝜃) = −7

25 with 𝜃 in IV, what is cos(𝜃)?

(−7

25)

2

+ cos2(𝜃) = 1 ⟹ cos2(𝜃) = 1 −49

625=

576

625⟹ cos(𝜃) = ±√

576

625= ±

24

25

𝐼𝑉 ⟹ cos(𝜃) > 0 ⟹ cos(𝜃) =24

25

22. cos(𝜃) =4

9 with 𝜃 in I, what is sin(𝜃)?

sin2(𝜃) + (4

9)

2

= 1 ⟹ sin2(𝜃) = 1 −16

81=

65

81⟹ sin(𝜃) = ±√

65

81= ±

√65

9

𝐼 ⟹ sin(𝜃) > 0 ⟹ sin(𝜃) =√65

9

23. sin(𝜃) =5

13 with 𝜃 in II, what is cos(𝜃)?

(5

13)

2

+ cos2(𝜃) = 1 ⟹ cos2(𝜃) = 1 −25

169=

144

169⟹ cos(𝜃) = ±√

144

169= ±

12

13

𝐼𝐼 ⟹ cos(𝜃) < 0 ⟹ cos(𝜃) = −12

13

24. cos(𝜃) = −2

11 with 𝜃 in III, what is sin(𝜃)?

sin2(𝜃) + (−2

11)

2

= 1 ⟹ sin2(𝜃) = 1 −4

121=

117

121⟹ sin(𝜃) = ±√

117

121= ±

√117

11

𝐼𝐼𝐼 ⟹ sin(𝜃) < 0 ⟹ sin(𝜃) = −√117

11

25. sin(𝜃) = −2

3 with 𝜃 in III, what is cos(𝜃)?

(−2

3)

2

+ cos2(𝜃) = 1 ⟹ cos2(𝜃) = 1 −4

9=

5

9⟹ cos(𝜃) = ±√

5

9= ±

√5

3

𝐼𝐼𝐼 ⟹ cos(𝜃) < 0 ⟹ cos(𝜃) = −√5

3

26. cos(𝜃) =28

53 with 𝜃 in IV, what is sin(𝜃)?

sin2(𝜃) + (28

53)

2

= 1 ⟹ sin2(𝜃) = 1 −784

2809=

2025

2809⟹ sin(𝜃) = ±

45

53

𝐼𝑉 ⟹ sin(𝜃) < 0 ⟹ sin(𝜃) = −45

53

27. sin(𝜃) =2√5

5 with

𝜋

2< 𝜃 < 𝜋, what is cos(𝜃)?

(2√5

5)

2

+ cos2(𝜃) = 1 ⟹ cos2(𝜃) = 1 −20

25=

5

25⟹ cos(𝜃) = ±

√5

5



𝜋

2< 𝜃 < 𝜋 ⟹ 𝐼𝐼 ⟹ cos(𝜃) < 0 ⟹ cos(𝜃) = −

√5

5

28. cos(𝜃) =√10

10 with 2𝜋 < 𝜃 <

5𝜋

2, what is sin(𝜃)?

sin2(𝜃) + (√10

10)

2

= 1 ⟹ sin2(𝜃) = 1 −10

100=

90

100⟹ sin(𝜃) = ±

3√10

10

2𝜋 < 𝜃 <5𝜋

2⟹ 𝐼 ⟹ sin(𝜃) > 0 ⟹ sin(𝜃) =

3√10

10

29. sin(𝜃) = −0.42 with 𝜋 < 𝜃 <3𝜋

2, what is cos(𝜃)?

(−0.42)2 + cos2(𝜃) = 1 ⟹ cos2(𝜃) = 1 − 0.1764 = 0.8236 ⟹ cos(𝜃) = ±√0.8236

𝜋 < 𝜃 <3𝜋

2⟹ 𝐼𝐼𝐼 ⟹ cos(𝜃) < 0 ⟹ cos(𝜃) = −√0.8236 ≈ −0.9075

30. cos(𝜃) = −0.98 with 𝜋

2< 𝜃 < 𝜋, what is sin(𝜃)?

sin2(𝜃) + (−0.98)2 = 1 ⟹ sin2(𝜃) = 1 − 0.9604 = 0.0396 ⟹ sin(𝜃) = ±√0.0396 𝜋

2< 𝜃 < 𝜋 ⟹ 𝐼𝐼 ⟹ sin(𝜃) > 0 ⟹ sin(𝜃) = √0.0396 ≈ 0.1990

Find all angles which satisfy the given equation. 𝜃 in degrees 𝜃 in radians sin(𝜃) cos(𝜃)

0∘ 0 0 1 30∘ 𝜋/6 1/2 √3/2 45∘ 𝜋/4 √2/2 √2/2 60∘ 𝜋/3 √3/2 1/2

90∘ 𝜋/2 1 0

31. sin(𝜃) =1

2⟹ reference angle of

𝜋

6 and in I or II ⟹ 𝜃 =

𝜋

6+ 2𝑛𝜋, 𝜃 =

5𝜋

6+ 2𝑛𝜋, 𝑛 ∈ ℤ

32. cos(𝜃) = −√3


𝜋

6 and in II or III ⟹ 𝜃 =

5𝜋

6+ 2𝑛𝜋, 𝜃 =

7𝜋

6+ 2𝑛𝜋,

𝑛 ∈ ℤ 33. sin(𝜃) = 0 ⟹ 𝜃 = 𝑛𝜋, 𝑛 ∈ ℤ

34. cos(𝜃) =√2


𝜋

4 and in I or IV ⟹ 𝜃 =

𝜋

4+ 2𝑛𝜋, 𝜃 =

7𝜋

4+ 2𝑛𝜋,

𝑛 ∈ ℤ

35. sin(𝜃) =√3


𝜋

3 and in I or II ⟹ 𝜃 =

𝜋

3+ 2𝑛𝜋, 𝜃 =

2𝜋

3+ 2𝑛𝜋,

𝑛 ∈ ℤ 36. cos(𝜃) = −1 ⟹ 𝜃 = 𝜋 + 2𝑛𝜋 = (2𝑛 + 1)𝜋

37. sin(𝜃) = −1 ⟹ 𝜃 =3𝜋

2+ 2𝑛𝜋, 𝑛 ∈ ℤ

38. cos(𝜃) =√3


𝜋

6 and in I or IV ⟹ 𝜃 =

𝜋

6+ 2𝑛𝜋, 𝜃 =

11𝜋

6+ 2𝑛𝜋,

𝑛 ∈ ℤ 39. cos(𝜃) = −1.001

|cos (𝜃)| ≤ 1 ⟹ no solution Solve the equation for 𝑡. (See comments following Theorem 10.5.) The distinction between 𝑡 as a real number and as an angle 𝜃 = 𝑡 radians is often blurred. We solve in exactly the same manner as we did above. Any properties of cosine and sine developed in this and following sections which regard them as functions of angles in radian measure apply equally well if the inputs are regarded as real numbers.



40. cos(𝑡) = 0 ⟹ quadrantal angle of 𝜋

2 or

3𝜋

2⟹ 𝑡 =

𝜋

2+ 2𝑛𝜋,

𝑡 =3𝜋

2+ 2𝑛𝜋 =

𝜋

2+ 𝜋 + 2𝑛𝜋 =

𝜋

2+ (2𝑛 + 1)𝜋, 𝑛 ∈ ℤ ⟹ 𝑡 =

𝜋

2+ 𝑛𝜋, 𝑛 ∈ ℤ

41. sin(𝑡) = −√2


𝜋

4 in II or III ⟹ 𝑡 =

3𝜋

4+ 2𝑛𝜋, 𝑡 =

5𝜋

4+ 2𝑛𝜋, 𝑛 ∈ ℤ

42. cos(𝑡) = 3, | cos(𝑡) ≤ 1 ⟹ no solution

43. sin(𝑡) = −1

2⟹reference angle of

𝜋

6 in III or IV ⟹ 𝑡 =

7𝜋

6+ 2𝑛𝜋, 𝑡 =

11𝜋

6+ 2𝑛𝜋, 𝑛 ∈ ℤ

44. cos(𝑡) =1


𝜋

3 in I or IV ⟹ 𝑡 =

𝜋

3+ 2𝑛𝜋, 𝑡 =

5𝜋

3+ 2𝑛𝜋, 𝑛 ∈ ℤ

45. sin(𝑡) = −2, |sin(𝑡)| ≤ 1 ⟹ no solution 46. cos(𝑡) = 1 ⟹ quadrantal angle of 0 ⟹ 𝑡 = 2𝑛𝜋, 𝑛 ∈ ℤ

47. sin(𝑡) = 1 ⟹ quadrantal angle of 𝜋

2⟹ 𝑡 =

𝜋

2+ 2𝑛𝜋, 𝑛 ∈ ℤ

48. cos(𝑡) = −√2

2⟹reference angle of

𝜋

4 in II or III ⟹ 𝑡 =

3𝜋

4+ 2𝑛𝜋, 𝑡 =

5𝜋

4+ 2𝑛𝜋, 𝑛 ∈ ℤ

Find the measurement of the missing angle and the lengths of the missing sides. Theorem 10.4: Suppose 𝜃 is an acute angle residing in a right triangle. If the length of the side adjacent to 𝜃 is 𝑎, the length of the side opposite 𝜃

is 𝑏, and the length of the hypotenuse is 𝑐, then cos(𝜃) =𝑎

𝑐 and

sin(𝜃) =𝑏

𝑐.

Pythagorean Theorem from geometry: 𝑎2 + 𝑏2 = 𝑐2 Euclidean geometry: The sum of the angles of a triangle is 180∘ or 𝜋 radians

55. Find 𝜃, 𝑏, 𝑐

90 + 30 + 𝜃 = 180 ⟹ 𝜃 = 60∘

cos(30∘) =1

𝑐=

√3

2⟹ 𝑐 =

2

√3

12 + 𝑏2 = 𝑐2 =4

3⟹ 𝑏2 =

1

3⟹ 𝑏 =

1

√3

56. Find 𝜃, 𝑎, 𝑐

𝜃 + 45 + 90 = 180 ⟹ 𝜃 = 45∘

cos(𝜃) = cos(45∘) =3

𝑐=

√2

2⟹ 𝑐 =

6

√2=

6√2

2= 3√2

𝑎2 + 32 = (3√2)2

⟹ 𝑎2 = 9 ⟹ 𝑎 = 3

57. Find 𝛼, 𝑎, 𝑏

𝛼 + 33 + 90 = 180 ⟹ 𝛼 = 57∘

sin(𝛼) =𝑎

8⟹ 𝑎 = 8 sin(57∘) ≈ 6.709

cos(𝛼) =𝑏

8⟹ 𝑏 = 8 cos(57∘) ≈ 4.357



58. Find 𝛽, 𝑎, 𝑐

𝛽 + 48 + 90 = 180 ⟹ 𝛽 = 42∘

cos(𝛽) =6

𝑐⟹ 𝑐 =

6

cos(42∘)≈ 8.074

sin(𝛽) =𝑎

𝑐⟹ 𝑎 = 𝑐 sin(𝛽) = (

6

cos(42∘)) sin(42∘)

≈ 5.402

Assume that 𝜃 is an acute angle in a right triangle and use thm 10.4 to find the requested side. Theorem 10.4: Suppose 𝜃 is an acute angle residing in a right triangle. If the length of the side adjacent to 𝜃 is 𝑎, the length of the side opposite 𝜃

is 𝑏, and the length of the hypotenuse is 𝑐, then cos(𝜃) =𝑎

𝑐 and

sin(𝜃) =𝑏

𝑐.

59. If 𝜃 = 12∘ and the side adjacent to 𝜃 has length 4, how long is the hypotenuse?

cos(𝜃) =𝑎

𝑐⟹ cos(12∘) =

4

𝑐⟹ 𝑐 =

4

cos(12∘)≈ 4.089

60. If 𝜃 = 78.123∘ and the hypotenuse has length 5280, how long is the side adjacent to 𝜃?

cos(𝜃) =𝑎

𝑐⟹ cos(78.123∘) =

𝑎

5280⟹ 𝑎 = 5280 cos(79.123∘) ≈ 1086.684

61. If 𝜃 = 59∘ and the side opposite 𝜃 has length 117.42, how long is the hypotenuse?

sin(𝜃) =𝑏

𝑐⟹ sin(59∘) =

117.42

𝑐⟹ 𝑐 =

117.42

sin(59∘)≈ 136.986

62. If 𝜃 = 5∘ and the hypotenuse has length 10, how long is the side opposite 𝜃?

sin(𝜃) =𝑏

𝑐⟹ sin(5∘) =

𝑏

10⟹ 𝑏 = 10 sin(5∘) ≈ 0.872

63. If 𝜃 = 5∘ and the hypotenuse has length 10, how long is the side adjacent to 𝜃?

cos(𝜃) =𝑎

𝑐⟹ cos(5∘) =

𝑎

10⟹ 𝑎 = 10 cos(5∘) ≈ 9.962

64. If 𝜃 = 37.5∘ and the side opposite 𝜃 has length 306, how long is the side adjacent to 𝜃?

sin(𝜃) =𝑏

𝑐⟹ sin(37.5∘) =

306

𝑐⟹ 𝑐 =

306

sin(37.5∘)

cos(𝜃) =𝑎

𝑐⟹ cos(37.5∘) =

𝑎 sin(37.5∘)

306⟹ 𝑎 =

306 cos(37.5∘)

sin(37.5∘)≈ 398.787

Let 𝜃 be the angle in standard position whose terminal side contains the given point then compute cos(𝜃) and sin(𝜃).

65. 𝑃(−7,24)

⟹ sin(𝜃) =24

25, cos(𝜃) = −

7

25

66. 𝑄(3,4)

⟹ sin(𝜃) =4

5, cos(𝜃) =

3

5



67. 𝑅(5, −9)

⟹ sin(𝜃) = −9

√106, cos(𝜃) =

5

√106

68. 𝑇(−2, −11)

⟹ sin(𝜃) = −11

5√5, cos(𝜃) = −

2

5√5

Find the equation of motion for the given scenario.

Assume that the center of the motion is the origin, the motion is counter-clockwise and that 𝑡 = 0 corresponds to a position along the positive 𝑥-axis. Equation 10.3: Suppose an object is traveling in a circular path of radius 𝑟 centered at the origin with constant angular velocity 𝜔. If 𝑡 = 0 corresponds to the point (𝑟, 0), then the 𝑥 and 𝑦 coordinates of the object are functions of 𝑡 and are given by 𝑥 = 𝑟 cos(𝜔𝑡) and 𝑦 =𝑟 sin(𝜔𝑡). Here, 𝜔 > 0 indicates a counter-clockwise direction and 𝜔 < 0 indicates a clockwise direction.

69. A point on the edge of the spinning yo-yo in exercise 50 from 10.1. Recall: The diameter of the yo-yo is 2.25 inches and it spins at 4500 rev/min.

𝜔 = (4500 rev

1 min) (

2𝜋 rad

1 rev) = 9000𝜋

rad

min

𝑟 =1

2(2.25 in) = 1.125 in

{𝑥 = 1.125 cos(9000𝜋𝑡)

𝑦 = 1.125 sin(9000𝜋𝑡) 𝑡 in minutes, 𝑥 and 𝑦 in inches

70. The yo-yo in exercise 52 from section 10.1. Recall: The radius of the circle is 28 inches and it completes one revolution in 3 seconds.

𝜔 = (1 rev

3 sec) (

2𝜋 rad

1 rev) =

2𝜋

3

rev

sec 𝑟 = 28 in

{𝑥 = 28 cos (

2𝜋𝑡

3)

𝑦 = 28 sin (2𝜋𝑡

3)

𝑡 in seconds, 𝑥 𝑎𝑛𝑑 𝑦 in inches

71. A point on the edge of the hard drive in exercise 53 from section 10.1. Recall: The diameter of the hard disk is 2.5 inches and it spins at 7200 RPM.

𝜔 = (7200 rev

1 min) (

2𝜋 rad

1 rev) = 14,400𝜋

rad

min

𝑟 =1

2(2.5 in) = 1.25 in

{𝑥 = 1.25 cos(14400𝜋𝑡)

𝑦 = 1.25 sin(14400𝜋𝑡) 𝑡 in minutes, 𝑥 and 𝑦 in inches

72. A passenger on the Big Wheel in exercise 55 from section 10.1. Recall: The diameter is 128 feet and completes 2 revolutions in 2 minutes, 7 seconds.

𝜔 = (2 rev

127 sec) (

2𝜋 rad

1 rev) =

4𝜋

127

rad

sec 𝑟 =

1

2(128 ft) = 64 ft

{𝑥 = 64 cos (

4𝜋𝑡

127)

𝑦 = 64 sin (4𝜋𝑡

127)

𝑡 in seconds, 𝑥 and 𝑦 in feet



73. Consider the numbers: 0, 1, 2, 3, 4. Take the square root of each of these numbers, then divide by 2. The resulting numbers should look hauntingly familiar. (See the values in the table on 722.)

𝑛 √𝑛 √𝑛/2 𝜃 sin(𝜃) cos(𝜃)

0 √0 = 0 0/2 = 0 0 0 1

1 √1 = 1 1/2 𝜋/6 1/2 √3/2 2 √2 √2/2 𝜋/4 √2/2 √2/2 3 √3 √3/2 𝜋/3 √3/2 1/2

4 √4 = 2 2/2 = 1 𝜋/2 1 0

74. Let 𝛼 and 𝛽 be the two acute angles of a right triangle.

(Thus 𝛼 and 𝛽 are complementary numbers.) Show that sin(𝛼) = cos(𝛽) and sin(𝛽) = cos(𝛼). The fact that co-functions of complementary angles are equal in this case is not an accident and a more general result will be given in section 10.4.

sin(𝛼) =𝑏

𝑐

sin(𝛽) =𝑎

𝑐

cos(𝛼) =𝑎

𝑐

cos(𝛽) =𝑏

𝑐

75. In the scenario of equation 10.3, we assumed that at 𝑡 = 0, the object was at the

point (𝑟, 0). If this is not the case, we can adjust the equations of motion by introducing a ‘time delay.’ If 𝑡0 > 0 is the first time the object passes through the point (𝑟, 0), show, with the help of your classmates, the equations of motion are 𝑥 =𝑟 cos(𝜔(𝑡 − 𝑡0)) and 𝑦 = 𝑟 sin(𝜔(𝑡 − 𝑡0)).

simple horizontal curve shift



10.3 Exercises

Use the given information to find the exact values of the remaining circular functions of 𝜃.

cos2(𝜃) + sin2(𝜃) = 1,

sec(𝜃) =1

cos(𝜃), csc(𝜃) =

1

sin(𝜃),

tan(𝜃) =sin(𝜃)

cos(𝜃)=

1

cot(𝜃), cot(𝜃) =

cos(𝜃)

sin(𝜃)=

1

tan(𝜃)

21. sin(𝜃) =3

5 with 𝜃 in II

(3

5)

2

+ cos2(𝜃) = 1 ⟹ cos2(𝜃) = 1 −9

25=

16

25⟹ cos(𝜃) = ±

4

5

𝐼𝐼 ⟹ cos(𝜃) < 0 ⟹ cos(𝜃) = −4

5

sec(𝜃) =1

−4/5= −

5

4, csc(𝜃) =

1

35

=5

3, tan(𝜃) =

3/5

−4/5= −

3

4, cot(𝜃) =

−4/5

3/5= −

4

3

22. tan(𝜃) =12

5 with 𝜃 in III

𝐼𝐼𝐼 ⟹ sin(𝜃) < 0, cos(𝜃) < 0 ⟹ sin(𝜃) = −12, cos(𝜃) = −5

cot(𝜃) =−5

−12=

5

12, sec(𝜃) =

1

−5= −

1

5, csc(𝜃) =

1

−12= −

1

12

23. csc(𝜃) =25

24 with 𝜃 in I

csc(𝜃) =1

sin(𝜃)⟹ sin(𝜃) =

24

25

(24

25)

2

+ cos2(𝜃) = 1 ⟹ cos2(𝜃) = 1 −576

625=

49

625⟹ cos(𝜃) = ±

7

25

𝐼 ⟹ cos(𝜃) =7

25

tan(𝜃) =24/25

7/25=

24

7 cot(𝜃) =

1

tan(𝜃)=

7

24 sec(𝜃) =

1

cos(𝜃)=

25

7

24. sec(𝜃) = 7 with 𝜃 in IV

sec(𝜃) =1

cos(𝜃)⟹ cos(𝜃) =

1

7

sin2(𝜃) + (1

7)

2

= 1 ⟹ sin2(𝜃) = 1 −1

49=

48

49⟹ sin(𝜃) = ±

4√3

7

𝐼𝑉 ⟹ sin(𝜃) < 0 ⟹ sin(𝜃) = −4√3

7

tan(𝜃) =−4√3/7

1/7= −4√3 cot(𝜃) =

1

tan(𝜃)= −

1

4√3 csc(𝜃) =

1

sin(𝜃)= −

7

4√3

25. csc(𝜃) = −10√91

91 with 𝜃 in III

csc(𝜃) =1

sin(𝜃)⟹ sin(𝜃) = −

91

10√91= −

√91

10



(−√91

10)

2

+ cos2(𝜃) = 1 ⟹ cos2(𝜃) = 1 −91

100=

9

100⟹ cos(𝜃) = ±

3

10

𝐼𝐼𝐼 ⟹ cos(𝜃) < 0 ⟹ cos(𝜃) = −3

10

tan(𝜃) =−√91/10

−3/10=

√91

3 cot(𝜃) =

1

tan(𝜃)=

3

√91 sec(𝜃) =

1

cos(𝜃)= −

10

3

26. cot(𝜃) = −23 with 𝜃 in II

𝐼𝐼 ⟹ sin(𝜃) > 0, cos(𝜃) < 0 ⟹ sin(𝜃) = 1, cos(𝜃) = −23

tan(𝜃) =1

cot(𝜃)= −

1

23 sec(𝜃) =

1

cos(𝜃)= −

1

23 csc(𝜃) =

1

sin(𝜃)= 1

27. tan(𝜃) = −2 with 𝜃 in IV

𝐼𝑉 ⟹ sin(𝜃) < 0, cos(𝜃) > 0 ⟹ sin(𝜃) = −2, cos(𝜃) = 1

cot(𝜃) =1

tan(𝜃)= −

1

2 sec(𝜃) =

1

cos(𝜃)= 1 csc(𝜃) =

1

sin(𝜃)= −

1

2

28. sec(𝜃) = −4 with 𝜃 in II

sec(𝜃) =1

cos(𝜃)⟹ cos(𝜃) = −

1

4

sin2(𝜃) + (−1

4)

2

= 1 ⟹ sin2(𝜃) = 1 −1

16=

15

16⟹ sin(𝜃) = ±

√15

4

𝐼𝐼 ⟹ sin(𝜃) > 0 ⟹ sin(𝜃) =√15

4


cos(𝜃)=

√15/4

−1/4= −√15 cot(𝜃) =

1

tan(𝜃)= −

1

√15 csc(𝜃) =

1

sin(𝜃)=

4

√15

29. cot(𝜃) = √5 with 𝜃 in III

𝐼𝐼𝐼 ⟹ sin(𝜃) < 0, cos(𝜃) < 0 ⟹ sin(𝜃) = −1, cos(𝜃) = −√5

tan(𝜃) =1

cot(𝜃)=

1

√5 sec(𝜃) =

1

cos(𝜃)= −

1

√5 csc(𝜃) =

1

sin(𝜃)= −1

30. cos(𝜃) =1

3 with 𝜃 in I

sin2(𝜃) + (1

3)

2

= 1 ⟹ sin2(𝜃) = 1 −1

9=

8

9⟹ sin(𝜃) = ±

2√2

3

𝐼 ⟹ sin(𝜃) > 0 ⟹ sin(𝜃) =2√2

3


cos(𝜃)=

2√2/3

1/3= 2√2 cot(𝜃) =

1

tan(𝜃)=

1

2√2=

√2

4

sec(𝜃) =1

cos(𝜃)= 3 csc(𝜃) =

1

sin(𝜃)=

3

2√2=

3√2

4



31. cot(𝜃) = 2 with 0 < 𝜃 <𝜋

2

0 < 𝜃 <𝜋

2⟹ 𝐼 ⟹ sin(𝜃) > 0, cos(𝜃) > 0 ⟹ sin(𝜃) = 1, cos(𝜃) = 2

cot(𝜃) =1

tan(𝜃)=

1

2 sec(𝜃) =

1

cos(𝜃)=

1

2 csc(𝜃) =

1

sin(𝜃)= 1

32. csc(𝜃) = 5 with 𝜋

2< 𝜃 < 𝜋

csc(𝜃) =1

sin(𝜃)⟹ sin(𝜃) =

1

5

(1

5)

2

+ cos2(𝜃) = 1 ⟹ cos2(𝜃) = 1 −1

25=

24

25⟹ cos(𝜃) = ±

2√6

5

𝜋

2< 𝜃 < 𝜋 ⟹ 𝐼𝐼 ⟹ cos(𝜃) < 0 ⟹ cos(𝜃) = −

2√6

5


cos(𝜃)=

1

−2√6 cot(𝜃) =

1

tan(𝜃)= −2√6 sec(𝜃) =

1

cos(𝜃)= −

5

2√6

33. tan(𝜃) = √10 with 𝜋 < 𝜃 <3𝜋

2

𝜋 < 𝜃 <3𝜋

2⟹ 𝐼𝐼𝐼 ⟹ sin(𝜃) < 0, cos(𝜃) < 0 ⟹ sin(𝜃) = −√10, cos(𝜃) = −1

cot(𝜃) =1

tan(𝜃)=

1

√10 sec(𝜃) =

1

cos(𝜃)= −1 csc(𝜃) =

1

sin(𝜃)= −

1

√10

34. sec(𝜃) = 2√5 with 3𝜋

2< 𝜃 < 2𝜋

sec(𝜃) =1

cos(𝜃)⟹ cos(𝜃) =

1

2√5=

√5

10

sin2(𝜃) + (1

2√5)

2

= 1 ⟹ sin2(𝜃) = 1 −1

20=

19

20⟹ sin(𝜃) = ±

√19

2√5= ±

√95

10

3𝜋

2< 𝜃 < 2𝜋 ⟹ 𝐼𝑉 ⟹ sin(𝜃) < 0 ⟹ sin(𝜃) = −

√95

10


cos(𝜃)= −

√95

√5= −√19 cot(𝜃) =

1

tan(𝜃)= −

1

√19 csc(𝜃) =

1

sin(𝜃)= −

10

√95

Find all of the angles which satisfy the equation.

𝜃 sin(𝜃) cos(𝜃) tan(𝜃) cot(𝜃) sec(𝜃) csc(𝜃)

0 0 1 0 𝐷𝑁𝐸 1 𝐷𝑁𝐸 𝜋/6 1/2 √3/2 1/√3 √3 2/√3 2

𝜋/4 √2/2 √2/2 1 1 √2 √2 𝜋/3 √3/2 1/2 √3 1/√3 2 2/√3 𝜋/2 1 0 𝐷𝑁𝐸 0 𝐷𝑁𝐸 1

43. tan(𝜃) = √3 ⟹ reference angle 𝜋

3 in I or III ⟹ 𝜃 =

𝜋

3+ 2𝑛𝜋, 𝜃 =

4𝜋

3+ 2𝑛𝜋, 𝑛 ∈ ℤ

44. sec(𝜃) = 2 ⟹ reference angle 𝜋

3 in I or IV ⟹ 𝜃 =

𝜋

3+ 2𝑛𝜋, 𝜃 =

5𝜋

3+ 2𝑛𝜋, 𝑛 ∈ ℤ

45. csc(𝜃) = −1 ⟹ quadrantal angle of 3𝜋

2⟹ 𝜃 =

3𝜋

2+ 2𝑛𝜋, 𝑛 ∈ ℤ



46. cot(𝜃) =√3

3⟹ reference angle

𝜋

6 in I or III ⟹ 𝜃 =

𝜋

6+ 2𝑛𝜋, 𝜃 =

7𝜋

6+ 2𝑛𝜋 =

𝜋

6+ 3𝑛𝜋

⟹𝜋

6+ 𝑛𝜋, 𝑛 ∈ ℤ

47. tan(𝜃) = 0 ⟹ quadrantal angles of 0 and 𝜋 ⟹ 𝜃 = 𝑛𝜋, 𝑛 ∈ ℤ 48. sec(𝜃) = 1 ⟹ quadrantal angle of 0 ⟹ 𝜃 = 2𝑛𝜋, 𝑛 ∈ ℤ

49. csc(𝜃) = 2 ⟹ reference angle of 𝜋

6 in I or II ⟹ 𝜃 =

𝜋

6+ 2𝑛𝜋, 𝜃 =

5𝜋

6+ 2𝑛𝜋, 𝑛 ∈ ℤ

50. cot(𝜃) = 0 ⟹ quadrantal angles of 𝜋

2 or

3𝜋

2⟹ 𝜃 =

𝜋

2+ 2𝑛𝜋, 𝜃 =

3𝜋

2+ 2𝑛𝜋 =

𝜋

2+ 3𝑛𝜋

⟹ 𝜃 =𝜋

2+ 𝑛𝜋, 𝑛 ∈ ℤ

51. tan(𝜃) = −1 ⟹ reference angle of 𝜋

4 in II or IV ⟹ 𝜃 =

3𝜋

4+ 2𝑛𝜋,

𝜃 =7𝜋

4+ 2𝑛𝜋 =

3𝜋

4+ 3𝑛𝜋 ⟹ 𝜃 =

3𝜋

4+ 𝑛𝜋, 𝑛 ∈ ℤ

52. sec(𝜃) = 0 ⟹1

cos(𝜃)= 0 ⟹ no solution

53. csc(𝜃) = −1

2⟹

1

sin(𝜃)= −

1

2⟹ sin(𝜃) = −2 ⟹ no solution

54. sec(𝜃) = −1 ⟹ quadrantal angle of 𝜋 ⟹ 𝜃 = 𝜋 + 2𝑛𝜋 = (2𝑛 + 1)𝜋, 𝑛 ∈ ℤ

55. tan(𝜃) = −√3 ⟹ reference angle 𝜋


2𝜋

3+ 2𝑛𝜋,

𝜃 =5𝜋

3+ 2𝑛𝜋 =

2𝜋

3+ 3𝑛𝜋 ⟹ 𝜃 =

2𝜋

3+ 𝑛𝜋, 𝑛 ∈ ℤ

56. csc(𝜃) = −2 ⟹ reference angle of 𝜋

6 in III or IV ⟹ 𝜃 =

7𝜋

6+ 2𝑛𝜋, 𝜃 =

11𝜋

6+ 2𝑛𝜋,

𝑛 ∈ ℤ

57. cot(𝜃) = −1 ⟹ reference angle of 𝜋


3𝜋

4+ 2𝑛𝜋,

𝜃 =7𝜋

4+ 2𝑛𝜋 =

3𝜋

4+ 3𝑛𝜋 ⟹ 𝜃 =

3𝜋

4+ 𝑛𝜋, 𝑛 ∈ ℤ

Solve the equation for 𝑡. Give exact values. 𝜃 sin(𝜃) cos(𝜃) tan(𝜃) cot(𝜃) sec(𝜃) csc(𝜃)

0 0 1 0 𝐷𝑁𝐸 1 𝐷𝑁𝐸 𝜋/6 1/2 √3/2 1/√3 √3 2/√3 2

𝜋/4 √2/2 √2/2 1 1 √2 √2 𝜋/3 √3/2 1/2 √3 1/√3 2 2/√3 𝜋/2 1 0 𝐷𝑁𝐸 0 𝐷𝑁𝐸 1

58. cot(𝑡) = 1 ⟹ reference angle of 𝜋

4 in I or III ⟹ 𝑡 =

𝜋

4+ 2𝑛𝜋, 𝑡 =

5𝜋

4+ 2𝑛𝜋, 𝑛 ∈ ℤ

59. tan(𝑡) =√3


𝜋


𝜋

6+ 2𝑛𝜋, 𝑡 =

7𝜋

6+ 2𝑛𝜋, 𝑛 ∈ ℤ

60. sec(𝑡) = −2√3


𝜋


𝜋

6+ 2𝑛𝜋, 𝑡 =

11𝜋

6+ 2𝑛𝜋,

𝑛 ∈ ℤ

61. csc(𝑡) = 0 ⟹ csc(𝑡) =1

sin(𝑡)≠ 0 ⟹ no solution

62. cot(𝑡) = −√3 ⟹ reference angle of 𝜋


𝜋

6+ 2𝑛𝜋, 𝑡 =

7𝜋

6+ 2𝑛𝜋, 𝑛 ∈ ℤ

63. tan(𝑡) = −√3


𝜋

6 in II or IV ⟹ 𝑡 =

5𝜋

6+ 2𝑛𝜋, 𝑡 =

11𝜋

6+ 2𝑛𝜋,

𝑛 ∈ ℤ

64. sec(𝑡) =2√3


𝜋


𝜋

6+ 2𝑛𝜋, 𝑡 =

11𝜋

6+ 2𝑛𝜋, 𝑛 ∈ ℤ



65. csc(𝑡) =2√3


𝜋

3 in I or II

⟹ 𝑡 =𝜋

3+ 2𝑛𝜋, 𝑡 =

2𝜋

3+ 2𝑛𝜋, 𝑛 ∈ ℤ

Use Theorem 10.10 to find the requested identities. Theorem 10.10. Suppose 𝜃 is an acute angle residing in a right triangle. If the length of the side adjacent to 𝜃 is 𝑎, the length of the side opposite 𝜃 is 𝑏, and the length of the hypotenuse is 𝑐, then

tan(𝜃) =𝑏

𝑎, sec(𝜃) =

𝑐

𝑎, csc(𝜃) =

𝑐

𝑏, cot(𝜃) =

𝑎

𝑏

Pythagorean Theorem: 𝑎2 + 𝑏2 = 𝑐2 Euclidean Geometry: The sum of the angles of a triangle is 180∘ or 𝜋 radians


90 + 60 + 𝜃 = 180 ⟹ 𝜃 = 30∘ =𝜋

6

sec (𝜋

6) =

𝑐

9⟹ 9 sec (

𝜋

6) = 𝑐 ⟹ 𝑐 =

18√3

3= 6√3

𝑎2 + 92 = (6√3)2

⟹ 𝑎2 + 81 = 108 ⟹ 𝑎 = 3√3

67. Find 𝛼, 𝑏, 𝑐

34 + 90 + 𝛼 = 180 ⟹ 𝛼 = 56∘

tan(𝛼) =12

𝑏⟹ 𝑏 = 12 cot(56∘) ≈ 8.094

122 + 𝑏2 = 𝑐2 ⟹ 144 + 144 cot2(56∘) = 𝑐2

⟹ 𝑐 ≈ 14.4746


𝜃 + 47 + 90 = 180 ⟹ 𝜃 = 43∘

tan(47∘) =6

𝑎⟹ 𝑎 = 6 cot(47∘) ≈ 5.595

𝑎2 + 62 = 𝑐2 ⟹ 𝑐2 = 36 cot2(47∘) + 36 ⟹ 𝑐 ≈ 8.204

69. Find 𝛽, 𝑏, 𝑐

𝛽 + 50 + 90 = 180 ⟹ 𝛽 = 40∘

tan(𝛽) =2.5

𝑏⟹ 𝑏 = 2.5 cot(40∘) ≈ 2.979

𝑏2 + (2.5)2 = 𝑐2 ⟹ 𝑐2 = 6.25 cot2(40∘) + 6.25

⟹ 𝑐 ≈ 3.889



Use Theorem 10.10 to answer the questions. Assume that 𝜃 is an angle in a right triangle. Theorem 10.10. Suppose 𝜃 is an acute angle residing in a right triangle. If the length of the side adjacent to 𝜃 is 𝑎, the length of the side opposite 𝜃 is 𝑏, and the length of the hypotenuse is 𝑐, then

tan(𝜃) =𝑏

𝑎, sec(𝜃) =

𝑐

𝑎, csc(𝜃) =

𝑐

𝑏, cot(𝜃) =

𝑎

𝑏

70. If 𝜃 = 30∘ and the side opposite 𝜃 has length 4, how long is the side adjacent to 𝜃?

tan(30∘) =4

𝑎⟹ 𝑎 = 4 cot(30∘) = 4√3

71. If 𝜃 = 15∘ and the hypotenuse has length 10, how long is the side opposite 𝜃?

csc(15∘) =10

𝑏⟹ 𝑏 = 10 sin(15∘)

72. If 𝜃 = 87∘ and the side adjacent to 𝜃 has length 2, how long is the side opposite 𝜃?

tan(87∘) =𝑏

2⟹ 𝑏 = 2 tan(87∘)

73. If 𝜃 = 38.2∘ and the side opposite 𝜃 has length 14, how long is the hypotenuse?

csc(38.2∘) =𝑐

14⟹ 𝑐 = 14 csc(38.2∘)

74. If 𝜃 = 2.05∘ and the hypotenuse has length 3.98, how long is the side adjacent to 𝜃?

sec(2.05) =3.98

𝑎⟹ 𝑎 = 3.98 cos(2.98∘)

75. If 𝜃 = 42∘ and the side adjacent to 𝜃 has length 31, how long is the side opposite 𝜃?

tan(42∘) =𝑏

31⟹ 𝑏 = 31 tan(31∘)

76. A tree standing vertically on level ground casts a 120 foot long shadow. The angle of elevation from the end of the shadow to the top of the tree is 21.4∘. Find the height of the tree to the nearest foot. With the help of your classmates, research the term umbra versa and see what it has to do with the shadow in this problem.

tan(21.4∘) =tree

120⟹ tree = 120 tan(21.4∘) ≈ 47 feet tall

77. The broadcast tower for radio station WSAZ has two

enormous flashing red lights on it: one at the very top and one a few feet below the top. From a point 5000 feet away from the base of the tower on level ground the angle of elevation to the top light is 7.970∘ and to the second light is 7.125∘. Find the distance between the lights to the nearest foot.

tan(7.125∘) =bottom light

5000⟹ bottom light ≈ 625 ft

tan(7.970∘) =top light

5000⟹ top light ≈ 700 ft

they are about 25 feet apart



78. On pg 753 we defined the angle of inclination (AKA the angle of elevation) and in this exercise we introduce a related angle – the angle of depression (AKA the angle of declination). The angle of depression of an object refers to the angle whose initial side is a horizontal line above the object and whose terminal side is the line-of-sight to the object below the horizontal.

a. Show that if the horizontal is above and parallel to level ground then the angle of depression (from observer to object) and the angle of inclination (from object to observer) will be congruent because they are alternate interior angles.

b. From a firetower 200 feet above level ground in the Sasquatch National Forest, a ranger spots a fire off in the distance. The angle of depression to the fire is 2.5∘. How far away from the base of the tower is the fire?

tan(2.5∘) =200

𝑑⟹ 𝑑 = 200 cot(2.5∘) ≈ 4581 ft

c. The ranger is part 78b sees a Sasquatch running directly from the fire towards the firetower. The ranger takes two sightings. At the first sighting, the angle of depression from the tower to the Sasquatch is 6∘. The second sighting, taken just 10 seconds later gives the angle of depression as 6.5∘. How fast is it running in miles per hour? Round your answer to the nearest mile per hour. If the Sasquatch keeps up this pace, how long will it take for the Sasquatch to reach the firetower from his location at the second sighting? Round your answer to the nearest minute.

tan(6.5∘) =200

𝑎⟹ 𝑎 = 200 cot(6.5∘) ≈ 1755.377

tan(6∘) =200

𝑏⟹ 𝑏 = 200 cot(6∘) ≈ 1902.873

distance traveled ≈ 147 feet

rate of travel = (147 ft

10 sec) (

1 mi

5280 ft) (

60 sec

1 min) (

60 min

1 hr) ≈ 10

mi

hr

time to tower = 1755 ft (1 mi

5280 ft) (

1 hr

10 mi) ≈ 0.03 hr ≈ 2 min

79. When I stand 30 feet away from a tree at home, the angle of elevation to the top of the tree is 50∘ and the angle of depression to the base of the tree is 10∘. What is the height of the tree? Round your answer to the nearest foot.

tan(10∘) =𝑎

30⟹ 𝑎 = 30 tan(10∘) ≈ 5.3 ft



tan(50∘) =𝑏

30⟹ 𝑏 = 30 tan(50∘) ≈ 35.8 ft

⟹ tree ≈ 41 ft 80. From the observation deck of the lighthouse at Sasquatch

Point 50 feet above the surface of Lake Ippizuti, a lifeguard spots a boat out on the lake sailing directly toward the lighthouse. The first sighting had an angle of depression of 8.2∘ and the second sighting had an angle of depression of 25.9∘. How far had the boat traveled between the sightings?

tan(8.2∘) =50

𝑎⟹ 𝑎 = 50 cot(8.2∘) ≈ 346.976 ft

tan(25.9∘) =50

𝑏⟹ 𝑏 = 50 cot(25.9∘) ≈ 102.971 ft

distance ≈ 244 ft 81. A guy wire 1000 feet long is attached to the top of a tower.

When pulled taught it makes a 43∘ angle with the ground. How tall is the tower? How far away from the base of the tower does the wire hit the ground?

sin(43∘) =tower

1000⟹ tower = 1000 sin(43∘) ≈ 682 ft

cos(43∘) =distance

1000⟹ distance = 1000 cos(43∘) ≈ 731 ft

Verify the identity. Assume that all quantities are defined. Note: when it comes to verifying identities, I choose to memorize as little as possible. Hence, these are completed based on the fewest identities needed overall rather than with an eye to the quickest way through that requires more formula memorization.

82. cos(𝜃) sec(𝜃) = 1

cos(𝜃) sec(𝜃) = cos(𝜃) (1

cos(𝜃)) sec(𝜃) =

1

cos(𝜃)

= 1

83. tan(𝜃) cos(𝜃) = sin(𝜃)

tan(𝜃) cos(𝜃) = (sin(𝜃)

cos(𝜃)) cos(𝜃) tan(𝜃) =

sin(𝜃)

cos(𝜃)

= sin(𝜃)

84. sin(𝜃) csc(𝜃) = 1

sin(𝜃) csc(𝜃) = sin(𝜃) (1

sin(𝜃)) csc(𝜃) =

1

sin(𝜃)

=sin(𝜃)

sin(𝜃)= 1

85. tan(𝜃) cot(𝜃) = 1

tan(𝜃) cot(𝜃) = (sin(𝜃)

cos(𝜃)) (

cos(𝜃)

sin(𝜃)) tan(𝜃) =

sin(𝜃)

cos(𝜃), cot(𝜃) =

cos(𝜃)

sin(𝜃)

= (sin(𝜃)

sin(𝜃)) (

cos(𝜃)

cos(𝜃)) = 1



86. csc(𝜃) cos(𝜃) = cot(𝜃)

csc(𝜃) cos(𝜃) = (1

sin(𝜃)) cos(𝜃) csc(𝜃) =

1

sin(𝜃)

=cos(𝜃)

sin(𝜃)= cot(𝜃) cot(𝜃) =

cos(𝜃)

sin(𝜃)

87. sin(𝜃)

cos2(𝜃)= sec(𝜃) tan(𝜃)

sec(𝜃) tan(𝜃) = (1

cos(𝜃)) (

sin(𝜃)

cos(𝜃)) sec(𝜃) =

1

cos(𝜃), tan(𝜃) =

sin(𝜃)

cos(𝜃)

=sin(𝜃)

cos2(𝜃)

88. cos(𝜃)

sin2(𝜃)= csc(𝜃) cot(𝜃)

csc(𝜃) cot(𝜃) = (1

sin(𝜃)) (

cos(𝜃)

sin(𝜃)) csc(𝜃) =

1

sin(𝜃), cot(𝜃) =

cos(𝜃)

sin(𝜃)

=cos(𝜃)

sin2(𝜃)

89. 1+sin(𝜃)

cos(𝜃)= sec(𝜃) + tan(𝜃)

sec(𝜃) + tan(𝜃) =1

cos(𝜃)+

sin(𝜃)

cos(𝜃) sec(𝜃) =

1


sin(𝜃)

cos(𝜃)

=1 + sin(𝜃)

cos(𝜃)

90. 1−cos(𝜃)

sin(𝜃)= csc(𝜃) − cot(𝜃)

csc(𝜃) − cot(𝜃) =1

sin(𝜃)−

cos(𝜃)

sin(𝜃) csc(𝜃) =

1


cos(𝜃)

sin(𝜃)

=1 − cos(𝜃)

sin(𝜃)

91. cos(𝜃)

1−sin2(𝜃)= sec(𝜃)

cos(𝜃)

1 − sin2(𝜃)=

cos(𝜃)

cos2(𝜃)

sin2(𝜃) + cos2(𝜃) = 1 ⟹ 1 − sin2(𝜃) = cos2(𝜃)

= (cos(𝜃)

cos(𝜃)) (

1

cos(𝜃))

=1

cos(𝜃)= sec(𝜃) sec(𝜃) =

1

cos(𝜃)

92. sin(𝜃)

1−cos2(𝜃)= csc(𝜃)

sin(𝜃)

1 − cos2(𝜃)=

sin(𝜃)

sin2(𝜃)

sin2(𝜃) + cos2(𝜃) = 1 ⟹ 1 − cos2(𝜃) = sin2(𝜃)



= (sin(𝜃)

sin(𝜃)) (

1

sin(𝜃))

=1

sin(𝜃)= csc(𝜃) csc(𝜃) =

1

sin(𝜃)

93. sec(𝜃)

1+tan2(𝜃)= cos(𝜃)

sec(𝜃)

1 + tan2(𝜃)=

sec(𝜃)

sec2(𝜃)

sin2(𝜃) + cos2(𝜃) = 1 ⟹ 1 + tan2(𝜃) = sec2(𝜃)

= (sec(𝜃)

sec(𝜃)) (

1

sec(𝜃))

=1

sec(𝜃)=

1

1/ cos(𝜃)= cos(𝜃) sec(𝜃) =

1

cos(𝜃)

94. csc(𝜃)

1+cot2(𝜃)= sin(𝜃)

csc(𝜃)

1 + cot2(𝜃)=

csc(𝜃)

csc2(𝜃)

sin2(𝜃) + cos2(𝜃) = 1 ⟹ 1 + cot2(𝜃) = csc2(𝜃)

=1

csc(𝜃)=

1

1/ sin(𝜃) csc(𝜃) =

1

sin(𝜃)

= sin(𝜃)

95. tan(𝜃)

sec2(𝜃)−1= cot(𝜃)

tan(𝜃)

sec2(𝜃) − 1=

tan(𝜃)

tan2(𝜃)

sin2(𝜃) + cos2(𝜃) = 1 ⟹ 1 + tan2(𝜃) = sec2(𝜃) ⟹ sec2(𝜃) − 1 = tan2(𝜃)

=1

tan(𝜃)=

1

sin(𝜃) / cos(𝜃) tan(𝜃) =

sin(𝜃)

cos(𝜃)

=cos(𝜃)

sin(𝜃)= cot(𝜃) cot(𝜃) =

cos(𝜃)

sin(𝜃)

96. cot(𝜃)

csc2(𝜃)−1= tan(𝜃)

cot(𝜃)

csc2(𝜃) − 1=

cot(𝜃)

cot2(𝜃)

sin2(𝜃) + cos2(𝜃) = 1 ⟹ 1 + cot2(𝜃) = csc2(𝜃) ⟹ csc2(𝜃) − 1 = cot2(𝜃)

=1

cot(𝜃)=

1

cos(𝜃) / sin(𝜃) cot(𝜃) =

cos(𝜃)

sin(𝜃)

=sin(𝜃)

cos(𝜃)= tan(𝜃) tan(𝜃) =

sin(𝜃)

cos(𝜃)

97. 4 cos2(𝜃) + 4 sin2(𝜃) = 4

4 cos2(𝜃) + 4 sin2(𝜃) = 4(cos2(𝜃) + sin2(𝜃))

= 4(1) = 4 sin2(𝜃) + cos2(𝜃) = 1

98. 9 − cos2(𝜃) − sin2(𝜃) = 8 9 − cos2(𝜃) − sin2(𝜃)



= 9 − (cos2(𝜃) + sin2(𝜃)) = 9 − 1 = 8 sin2(𝜃) + cos2(𝜃) = 1

99. tan3(𝜃) = tan(𝜃) sec2(𝜃) − tan(𝜃)

tan3(𝜃) = tan(𝜃) tan2(𝜃) = tan(𝜃) (sec2(𝜃) − 1) sin2(𝜃) + cos2(𝜃) = 1

⟹ tan2(𝜃) + 1 = sec2(𝜃) ⟹ tan2(𝜃) = sec2(𝜃) − 1

= tan(𝜃) sec2(𝜃) − tan(𝜃)

100. sin5(𝜃) = (1 − cos2(𝜃))2 sin(𝜃) sin5(𝜃) = (sin2(𝜃))2 sin(𝜃)

= (1 − cos2(𝜃))2 sin(𝜃) sin2(𝜃) + cos2(𝜃) = 1 ⟹ sin2(𝜃) = 1 − cos2(𝜃)

101. sec10(𝜃) = (1 + tan2(𝜃))4 sec2(𝜃)

sec10(𝜃) = (sec2(𝜃))4 sec2(𝜃) = (1 + tan2(𝜃))4 sec2(𝜃) sin2(𝜃) + cos2(𝜃) = 1

⟹ tan2(𝜃) + 1 = sec2(𝜃)

102. cos2(𝜃) tan3(𝜃) = tan(𝜃) − sin(𝜃) cos(𝜃)

cos2(𝜃) tan3(𝜃) = cos2(𝜃) (sin3(𝜃)

cos3(𝜃)) tan(𝜃) =

sin(𝜃)

cos(𝜃)

=sin3(𝜃)

cos(𝜃)=

sin2(𝜃) sin(𝜃)

cos(𝜃)=

(1 − cos2(𝜃)) sin(𝜃)

cos(𝜃)

sin2(𝜃) + cos2(𝜃) = 1 ⟹ sin2(𝜃) = 1 − cos2(𝜃)

=sin(𝜃)

cos(𝜃)−

sin(𝜃) cos2(𝜃)

cos(𝜃)= tan(𝜃) − sin(𝜃) cos(𝜃) tan(𝜃) =

sin(𝜃)

cos(𝜃)

103. sec4(𝜃) − sec2(𝜃) = tan2(𝜃) + tan4(𝜃)

sec4(𝜃) − sec2(𝜃) = sec2(𝜃) (sec2(𝜃) − 1) = (tan2(𝜃) + 1)(sec2(𝜃) − 1) sin2(𝜃) + cos2(𝜃) = 1

⟹ tan2(𝜃) + 1 = sec2(𝜃) = (tan2(𝜃) + 1) tan2(𝜃) = tan4(𝜃) + tan2(𝜃) ⟹ tan2(𝜃) = sec2(𝜃) − 1

104. cos(𝜃)+1

cos(𝜃)−1=

1+sec(𝜃)

1−sec(𝜃)

(cos(𝜃) + 1

cos(𝜃) − 1) (

1/ cos(𝜃)

1/ cos(𝜃)) =

1 + 1/ cos(𝜃)

1 − 1/ cos(𝜃)

=1 + sec(𝜃)

1 − sec(𝜃) sec(𝜃) =

1

cos(𝜃)

105. sin(𝜃)+1

sin(𝜃)−1=

1+csc(𝜃)

1−csc(𝜃)

(sin(𝜃) + 1

sin(𝜃) − 1) (

1/ sin(𝜃)

1/ sin(𝜃)) =

1 + 1/ sin(𝜃)

1 − 1/ sin(𝜃)

=1 + csc(𝜃)

1 − csc(𝜃) csc(𝜃) =

1

sin(𝜃)



106. 1−cot(𝜃)

1+cot(𝜃)=

tan(𝜃)−1

tan(𝜃)+1

1 − cot(𝜃)

1 + cot(𝜃)=

1 −1

tan(𝜃)

1 +1

tan(𝜃)

cot(𝜃) =

1

tan(𝜃)

=

tan(𝜃) − 1tan(𝜃)

tan(𝜃) + 1tan(𝜃)

= (tan(𝜃) − 1

tan(𝜃)) (

tan(𝜃)

tan(𝜃) + 1)

=tan(𝜃) − 1

tan(𝜃) + 1

107. 1−tan(𝜃)

1+tan(𝜃)=

cos(𝜃)−sin(𝜃)

cos(𝜃)+sin(𝜃)

1 − tan(𝜃)

1 + tan(𝜃)=

1 −sin(𝜃)cos(𝜃)

1 +sin(𝜃)cos(𝜃)

tan(𝜃) =

sin(𝜃)

cos(𝜃)

=

cos(𝜃) − sin(𝜃)cos(𝜃)

cos(𝜃) + sin(𝜃)cos(𝜃)

= (cos(𝜃) − sin(𝜃)

cos(𝜃)) (

cos(𝜃)

cos(𝜃) + sin(𝜃))

=cos(𝜃) − sin(𝜃)

cos(𝜃) + sin(𝜃)

108. tan(𝜃) + cot(𝜃) = sec(𝜃) csc(𝜃)

tan(𝜃) + cot(𝜃) =sin(𝜃)

cos(𝜃)+

cos(𝜃)

sin(𝜃) tan(𝜃) =

sin(𝜃)


cos(𝜃)

sin(𝜃)

=(sin2(𝜃) + cos2(𝜃))

cos(𝜃) sin(𝜃)=

1

cos(𝜃) sin(𝜃)

sin2(𝜃) + cos2(𝜃) = 1

= (1

cos(𝜃)) (

1

sin(𝜃)) = sec(𝜃) csc(𝜃) sec(𝜃) =

1

cos(𝜃), csc(𝜃) =

1

sin(𝜃)

109. csc(𝜃) − sin(𝜃) = cot(𝜃) cos(𝜃)

csc(𝜃) − sin(𝜃) =1

sin(𝜃)− sin(𝜃) csc(𝜃) =

1

sin(𝜃)

=1 − sin2(𝜃)

sin(𝜃)=

cos2(𝜃)

sin(𝜃)

sin2(𝜃) + cos2(𝜃) = 1 ⟹ cos2(𝜃) = 1 − sin2(𝜃)

= (cos(𝜃)

sin(𝜃)) cos(𝜃) = cot(𝜃) cos(𝜃) cot(𝜃) =

cos(𝜃)

sin(𝜃)

110. cos(𝜃) − sec(𝜃) = − tan(𝜃) sin(𝜃)

cos(𝜃) − sec(𝜃) = cos(𝜃) −1

cos(𝜃) sec(𝜃) =

1

cos(𝜃)

=cos2(𝜃) − 1

cos(𝜃)= −

sin2(𝜃)

cos(𝜃)


⟹ − sin2(𝜃) = cos2(𝜃) − 1



= − (sin(𝜃)

cos(𝜃)) sin(𝜃) = − tan(𝜃) sin(𝜃) tan(𝜃) =

sin(𝜃)

cos(𝜃)

111. cos(𝜃) (tan(𝜃) + cot(𝜃)) = csc(𝜃)

cos(𝜃) (tan(𝜃) + cot(𝜃))

= cos(𝜃) (sin(𝜃)

cos(𝜃)+

cos(𝜃)

sin(𝜃))



cos(𝜃)

sin(𝜃)

= sin(𝜃) +cos2(𝜃)

sin(𝜃)=

sin2(𝜃) + cos2(𝜃)

sin(𝜃)=

1

sin(𝜃)

sin2(𝜃) + cos2(𝜃) = 1

= csc(𝜃) csc(𝜃) =

1

sin(𝜃)

112. sin(𝜃) (tan(𝜃) + cot(𝜃)) = sec(𝜃)

sin(𝜃) (tan(𝜃) + cot(𝜃))

= sin(𝜃) (sin(𝜃)

cos(𝜃)+

cos(𝜃)

sin(𝜃))



cos(𝜃)

sin(𝜃)

=sin2(𝜃)

cos(𝜃)+ cos(𝜃) =

sin2(𝜃) + cos2(𝜃)

cos(𝜃)=

1

cos(𝜃)

sin2(𝜃) + cos2(𝜃) = 1

= sec(𝜃) sec(𝜃) =

1

cos(𝜃)

113. 1

1−cos(𝜃)+

1

1+cos(𝜃)= 2 csc2(𝜃)

1

1 − cos(𝜃)+

1

1 + cos(𝜃)=

1 + cos(𝜃) + 1 − cos(𝜃)

(1 − cos(𝜃))(1 + cos(𝜃))

=2

1 − cos2(𝜃)=

2

sin2(𝜃)


= 2 csc2(𝜃) csc(𝜃) =

1

sin(𝜃)

114. 1

sec(𝜃)+1+

1

sec(𝜃)−1= 2 csc(𝜃) cot(𝜃)

1

sec(𝜃) + 1+

1

sec(𝜃) − 1=

sec(𝜃) + 1 + sec(𝜃) − 1

(sec(𝜃) + 1)(sec(𝜃) − 1)

=2 sec(𝜃)

sec2(𝜃) − 1=

2 sec(𝜃)

tan2(𝜃)

sin2(𝜃) + cos2(𝜃) = 1 ⟹ tan2(𝜃) + 1 = sec2(𝜃) ⟹ tan2(𝜃) = sec2(𝜃) − 1

= 2 (1

cos(𝜃)) (

cos2(𝜃)

sin2(𝜃)) sec(𝜃) =

1


sin(𝜃)

cos(𝜃)

=2 cos(𝜃)

sin2(𝜃)= 2 (

1

sin(𝜃)) (

cos(𝜃)

sin(𝜃))

= 2 csc(𝜃) cot(𝜃) csc(𝜃) =

1


cos(𝜃)

sin(𝜃)

115. 1

csc(𝜃)+1+

1

csc(𝜃)−1= 2 sec(𝜃) tan(𝜃)

1

csc(𝜃) + 1+

1

csc(𝜃) − 1=

csc(𝜃) + 1 + csc(𝜃) − 1

(csc(𝜃) + 1)(csc(𝜃) − 1)



=2 csc(𝜃)

csc2(𝜃) − 1=

2 csc(𝜃)

cot2(𝜃)

sin2(𝜃) + cos2(𝜃) = 1 ⟹ cot2(𝜃) + 1 = csc2(𝜃) ⟹ cot2(𝜃) = csc2(𝜃) − 1

= 2 (1

sin(𝜃)) (

sin2(𝜃)

cos2(𝜃)) csc(𝜃) =

1


cos(𝜃)

sin(𝜃)

=2 sin(𝜃)

cos2(𝜃)= 2 (

1

cos(𝜃)) (

sin(𝜃)

cos(𝜃))

= 2 sec(𝜃) tan(𝜃) sec(𝜃) =

1


sin(𝜃)

cos(𝜃)

116. 1

csc(𝜃)−cot(𝜃)−

1

csc(𝜃)+cot(𝜃)= 2 cot(𝜃)

1

csc(𝜃) − cot(𝜃)−

1

csc(𝜃) + cot(𝜃)

=(csc(𝜃) + cot(𝜃)) − (csc(𝜃) − cot(𝜃))

csc2(𝜃) − cot2(𝜃)

=2 cot(𝜃)

csc2(𝜃) − cot2(𝜃)= 2 cot(𝜃)

sin2(𝜃) + cos2(𝜃) = 1 ⟹ cot2(𝜃) + 1 = csc2(𝜃) ⟹ 1 = csc2(𝜃) − cot2(𝜃)

117. cos(𝜃)

1−tan(𝜃)+

sin(𝜃)

1−cot(𝜃)= sin(𝜃) + cos(𝜃)

cos(𝜃)

1 − tan(𝜃)+

sin(𝜃)

1 − cot(𝜃)

=cos(𝜃)

1 −sin(𝜃)cos(𝜃)

+sin(𝜃)

1 −cos(𝜃)sin(𝜃)



cos(𝜃)

sin(𝜃)

=

cos(𝜃)cos(𝜃) − sin(𝜃)

cos(𝜃)+

sin(𝜃)sin(𝜃) − cos(𝜃)

sin(𝜃)

=cos2(𝜃)

cos(𝜃) − sin(𝜃)−

sin2(𝜃)

cos(𝜃) − sin(𝜃)

=cos2(𝜃) − sin2(𝜃)

cos(𝜃) − sin(𝜃)=

(cos(𝜃) − sin(𝜃))(cos(𝜃) + sin(𝜃))

cos(𝜃) − sin(𝜃)= sin(𝜃) + cos(𝜃)

118. 1

sec(𝜃)+tan(𝜃)= sec(𝜃) − tan(𝜃)

(1

sec(𝜃) + tan(𝜃)) (

sec(𝜃) − tan(𝜃)

sec(𝜃) − tan(𝜃))

=sec(𝜃) − tan(𝜃)

sec2(𝜃) − tan2(𝜃)= sec(𝜃) − tan(𝜃)

sin2(𝜃) + cos2(𝜃) = 1 ⟹ tan2(𝜃) + 1 = sec2(𝜃) ⟹ 1 = sec2(𝜃) − tan2(𝜃)

119. 1

sec(𝜃)−tan(𝜃)= sec(𝜃) + tan(𝜃)

(1

sec(𝜃) − tan(𝜃)) (

sec(𝜃) + tan(𝜃)

sec(𝜃) + tan(𝜃))

=sec(𝜃) + tan(𝜃)

sec2(𝜃) − tan2(𝜃)= sec(𝜃) + tan(𝜃)

sin2(𝜃) + cos2(𝜃) = 1 ⟹ tan2(𝜃) + 1 = sec2(𝜃) ⟹ 1 = sec2(𝜃) − tan2(𝜃)



120. 1

csc(𝜃)−cot(𝜃)= csc(𝜃) + cot(𝜃)

(1

csc(𝜃) − cot(𝜃)) (

csc(𝜃) + cot(𝜃)

csc(𝜃) + cot(𝜃))

=csc(𝜃) + cot(𝜃)

csc2(𝜃) − cot2(𝜃)= csc(𝜃) + cot(𝜃)


121. 1

csc(𝜃)+cot(𝜃)= csc(𝜃) − cot(𝜃)

(1

csc(𝜃) + cot(𝜃)) (

csc(𝜃) − cot(𝜃)

csc(𝜃) − cot(𝜃))

=csc(𝜃) − cot(𝜃)

csc2(𝜃) − cot2(𝜃)= csc(𝜃) − cot(𝜃)


122. 1

1−sin(𝜃)= sec2(𝜃) + sec(𝜃) tan(𝜃)

sec2(𝜃) + sec(𝜃) tan(𝜃)

=1

cos2(𝜃)+ (

1

cos(𝜃)) (

sin(𝜃)

cos(𝜃))

sec(𝜃) =1


sin(𝜃)

cos(𝜃)

=1 + sin(𝜃)

cos2(𝜃)=

1 + sin(𝜃)

1 − sin2(𝜃)


=1 + sin(𝜃)

(1 + sin(𝜃))(1 − sin(𝜃))=

1

1 − sin(𝜃)

123. 1

1+sin(𝜃)= sec2(𝜃) − sec(𝜃) tan(𝜃)

sec2(𝜃) − sec(𝜃) tan(𝜃)

=1

cos2(𝜃)− (

1

cos(𝜃)) (

sin(𝜃)

cos(𝜃))

sec(𝜃) =1


sin(𝜃)

cos(𝜃)

=1 − sin(𝜃)

cos2(𝜃)=

1 − sin(𝜃)

1 − sin2(𝜃)


=1 − sin(𝜃)

(1 + sin(𝜃))(1 − sin(𝜃))=

1

1 + sin(𝜃)

124. 1

1−cos(𝜃)= csc2(𝜃) + csc(𝜃) cot(𝜃)

csc2(𝜃) + csc(𝜃) cot(𝜃)

=1

sin2(𝜃)+ (

1

sin(𝜃)) (

cos(𝜃)

sin(𝜃))

csc(𝜃) =1


cos(𝜃)

sin(𝜃)

=1 + cos(𝜃)

sin2(𝜃)=

1 + cos(𝜃)

1 − cos2(𝜃)


=1 + cos(𝜃)

(1 + cos(𝜃))(1 − cos(𝜃))=

1

1 − cos(𝜃)



125. 1

1+cos(𝜃)= csc2(𝜃) − csc(𝜃) cot(𝜃)

csc2(𝜃) − csc(𝜃) cot(𝜃)

=1

sin2(𝜃)− (

1

sin(𝜃)) (

cos(𝜃)

sin(𝜃))

csc(𝜃) =1


cos(𝜃)

sin(𝜃)

=1 − cos(𝜃)

sin2(𝜃)=

1 − cos(𝜃)

1 − cos2(𝜃)


=1 − cos(𝜃)

(1 + cos(𝜃))(1 − cos(𝜃))=

1

1 + cos(𝜃)

126. cos(𝜃)

1+sin(𝜃)=

1−sin(𝜃)

cos(𝜃)

cos(𝜃)

1 + sin(𝜃)= (

cos(𝜃)

1 + sin(𝜃)) (

1 − sin(𝜃)

1 − sin(𝜃))

=cos(𝜃) (1 − sin(𝜃))

1 − sin2(𝜃)=

cos(𝜃) (1 − sin(𝜃))

cos2(𝜃)


=1 − sin(𝜃)

cos(𝜃)

127. csc(𝜃) − cot(𝜃) =sin(𝜃)

1+cos(𝜃)

csc(𝜃) − cot(𝜃) =1

sin(𝜃)−

cos(𝜃)

sin(𝜃) csc(𝜃) =

1


cos(𝜃)

sin(𝜃)

=1 − cos(𝜃)

sin(𝜃)(

1 + cos(𝜃)

1 + cos(𝜃))

=1 − cos2(𝜃)

sin(𝜃) (1 + cos(𝜃))=

sin2(𝜃)

sin(𝜃) (1 + cos(𝜃))


=sin(𝜃)

(1 + cos(𝜃))

128. 1−sin(𝜃)

1+sin(𝜃)= (sec(𝜃) − tan(𝜃))2

1 − sin(𝜃)

1 + sin(𝜃)= (

1 − sin(𝜃)

1 + sin(𝜃)) (

1 − sin(𝜃)

1 − sin(𝜃))

=(1 − sin(𝜃))2

1 − sin2(𝜃)=

(1 − sin(𝜃))2

cos2(𝜃)


= (1 − sin(𝜃)

cos(𝜃))

2

= (1

cos(𝜃)−

sin(𝜃)

cos(𝜃))

2

= (sec(𝜃) − tan(𝜃))2 sec(𝜃) =

1


sin(𝜃)

cos(𝜃)

Verify the identity. 129. ln|𝑠𝑒𝑐(𝜃)| = − ln|𝑐𝑜𝑠(𝜃)|

ln|𝑠𝑒𝑐(𝜃)| = ln |1

𝑐𝑜𝑠(𝜃)| sec(𝜃) =

1

cos(𝜃)

= ln|1| − ln|𝑐𝑜𝑠(𝜃)| ln (𝑎

𝑏) = ln(𝑎) − ln(𝑏)

= − ln|𝑐𝑜𝑠(𝜃)| ln(1) = 0



⟹ area =1

2(base)(altitude) =

1

2sin(𝜃)

b. Show that the circular sector 𝑂𝑃𝐵 with central angle 𝜃 has area 1

2𝜃

from 10.1 #56, the area of a circular sector is given by

𝐴 =1

2𝑟2𝜃 =

1

2(1)2𝜃 =

1

2𝜃

c. Show that the triangle 𝑂𝑄𝐵 has area 1

2tan(𝜃)

tan(𝜃) =𝑎𝑙𝑡𝑖𝑡𝑢𝑑𝑒

𝑏𝑎𝑠𝑒=

𝑎𝑙𝑡𝑖𝑡𝑢𝑑𝑒

1= 𝑎𝑙𝑡𝑖𝑡𝑢𝑑𝑒

𝑎𝑟𝑒𝑎 =1

2(𝑏𝑎𝑠𝑒)(𝑎𝑙𝑡𝑖𝑡𝑢𝑑𝑒) =

1

2(1)(tan(𝜃)) =

1

2tan(𝜃)

d. Comparing areas, show that sin(𝜃) < 𝜃 < tan(𝜃) for 0 < 𝜃 <𝜋

2

𝑎𝑟𝑒𝑎 𝑜𝑓 𝑠𝑚𝑎𝑙𝑙 𝑡𝑟𝑖𝑎𝑛𝑔𝑙𝑒 < 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑐𝑖𝑟𝑐𝑢𝑙𝑎𝑟 𝑠𝑒𝑐𝑡𝑜𝑟< 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑏𝑖𝑔 𝑡𝑟𝑖𝑎𝑛𝑔𝑙𝑒

⟹1

2sin(𝜃) <

1

2𝜃 <

1

2tan(𝜃) ⟹ sin(𝜃) < 𝜃 < tan(𝜃)

angle restriction comes from the picture representing only angles within quadrant I

e. Use the inequality sin(𝜃) < 𝜃 to show that sin(𝜃)

𝜃< 1 for 0 < 𝜃 <

𝜋

2

f. Use the inequality 𝜃 < tan(𝜃) to show that cos(𝜃) <sin(𝜃)

𝜃 for 0 < 𝜃 <

𝜋

2.

Combine this with the previous part to complete the proof.

10.1 exercises - elgin - instructor... · exercises 10.1.2 instructor use only k. campbell...

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